Question

Use the pumping lemma to show that the following languages are not regular

a. Ai = n 0 is not regular.
b. A2 = {www | w ∈ {a, b)*)
c.A₃+= j =i or j 2i

Answer 1

Final Answer:

a. The language A₁ = i ≥ 0 is not regular.

b. The language A₂ = {www | w ∈ {a, b}*} is not regular.

c. The language A₃₊ = aᶦbʲ is not regular.

Explanation:

To prove the non-regularity of A₁, we apply the Pumping Lemma for Regular Languages. Assume A₁ is regular, and let p be the pumping length. Consider the string s = 0ᵖ1². According to the Pumping Lemma, s can be divided into three parts, xyz, satisfying the conditions. However, no matter how we pump y, the resulting string violates the condition that the number of 0s and the square of the number of 1s are not equal. Thus, A₁ cannot be regular.

For A₂, assume it is regular. Let p be the pumping length. Consider the string s = aᵖbaᵖbaᵖb. According to the Pumping Lemma, s can be divided into xyz. By pumping y, the resulting string will have a different number of 'a's in each part, violating the condition for A₂. Therefore, A₂ is not regular.

Now, for A₃₊, assume it is regular with pumping length p. Consider the string s = aᵖb²a²b². According to the Pumping Lemma, s can be divided into xyz. Pumping y multiple times will lead to different values for i and j, violating the condition for A₃₊. Hence, A₃₊ is not regular.

In all cases, we have shown that the given languages are not regular based on the Pumping Lemma, demonstrating their non-regularity.

Answer 2

a. The language Ai = n ≥ 0 is not regular, as it violates the pumping lemma for regular languages. b. The language A2 = {www | w ∈ {a, b}*} is not regular, as it also violates the pumping lemma. c. The language A₃+ = j = i or j > 2i is not regular, as it violates the pumping lemma conditions for regularity.

Step-by-step explanation:

The pumping lemma for regular languages states that for any regular language L, there exists a constant p (the pumping length) such that any string s in L with length at least p can be divided into three parts, xyz, satisfying certain conditions. These conditions are not met for languages Ai, A2, and A₃+.

For Ai, the requirement of having equal numbers of 0s, 1s, and 2s makes it impossible to satisfy the conditions of the pumping lemma. A2 violates the lemma because the repeated segments cannot be pumped in a way that maintains equality. A₃+ fails the pumping lemma as the conditions for the lengths of a and b cannot be consistently maintained when pumped.